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Is there a way of proving the existence of antiderivatives (of continuous functions on a compact subset of the real line) without using tools of integration?

This is an exercise in: last page. Apparently one should somehow be able to use the Weierstrass theorem.

A related question is now of course: can we prove the fundamental theorem of calculus using Weierstrass' theorem?

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maybe by using series expansion and "antideriving" term by term (wild guess) – GPerez Feb 5 '14 at 15:37
This is integration. what do you mean by "using tools of"? – Robert Israel Feb 5 '14 at 15:40
There's probably a reason it's called the fundamental theorem of calculus. – dls Feb 5 '14 at 16:52
Basically without using the fundamental theorem. Maybe the argument could run like this: it is obvious for polynomials; so let $f\in C^0[a,b]$ and $p_k$ a sequence of polynomials uniformly converging against $f$. If we can argue that we have a sequence of antiderivatives for the polynomials, then they converge somehow in the sup norm (bc cauchy sequence?). Its limit is then $C^1$ and its derivative is $f$. – user88576 Feb 5 '14 at 22:20
@amoreacceptablename: Yes, that should work. Note that if polynomials $p$ and $q$ have antiderivatives $P$ and $Q$ with $P(a) = Q(a)$, and $|p - q| \le \epsilon$ on an interval $[a,b]$, then $|P - Q| \le \epsilon |b - a|$ on that interval. – Robert Israel Feb 6 '14 at 2:49

I propose now the following proof (thanks to R. Israel for the hint): One can easily write down antiderivatives for polynomials. Let now $f\in C^0[a,b]$ and $(p_k)$ be a sequence of polynomials with $p_k\rightarrow f$ uniformly (by the Weierstrass approximation theorem); let $(P_k)$ be a sequence of polynomials with $P'_k=p_k$, made unique with this property by requiring that $P_k(a)$ be equal to some constant $C$ (independent of $k$) for all $k$. Let $\varepsilon>0$; since $(p_k)$ is a Cauchy sequence there is an index $N$ such that $||p_m-p_n||<\varepsilon/|b-a|$ for all $m,n\geq N$ (here $||\cdot||$ denotes the sup norm). By the Mean value theorem we have for all $x\in [a,b]$ and indices $n,m$ $|P_m(x)-P_n(x)|=|P_m(x)-P_n(x)-(P_m(a)-P_n(a)|\leq||p_m-p_n|||x-a|\leq||p_m-p_n|||b-a|$, hence, taking the sup wrt $x\in[a,b]$ gives $||P_m-P_n||\leq||p_m-p_n|||b-a|$. Therefore $(P_k)$ is a Cauchy sequence as $||P_m-P_n||<\varepsilon|b-a|/|b-a|$ whenever $m,n\geq N$; let $F\in C^0[a,b]$ be its limit (exists by the completeness of $C^0[a,b]$ wrt the sup norm). Then $F\in C^1(a,b)$ and $F'=f$, completing the proof.

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Feel free to comment and give corrections or better arguments – user88576 Feb 6 '14 at 13:48

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